第四届 山东省ACM Rescue The Princess（计算几何）

Rescue ThePrincess

Time Limit:1000ms   Memory limit: 65536K  有疑问？点这里^_^

题目描述

    Several daysago, a beast caught a beautiful princess and the princess was put in prison. Torescue the princess, a prince who wanted to marry the princess set outimmediately. Yet, the beast set a maze. Only if the prince find out the maze’sexit can he save the princess.

    Now, here comes theproblem. The maze is a dimensional plane. The beast is smart, and he hidden theprincess snugly. He marked two coordinates of an equilateral triangle in the maze. The two markedcoordinates are A(x₁,y₁) and B(x₂,y₂).The third coordinate C(x₃,y₃) is the maze’s exit. If theprince can find out the exit, he can save the princess. After the prince comesinto the maze, he finds out the A(x₁,y₁) and B(x₂,y₂),but he doesn’t know where the C(x₃,y₃) is. The princeneed your help. Can you calculate the C(x₃,y₃) and tell him?

输入

    The first line is aninteger T(1 <= T <= 100) which is the number of test cases. T test casesfollow. Each test case contains two coordinates A(x₁,y₁)and B(x₂,y₂), described by four floating-point numbers x₁,y₁, x₂, y₂( |x₁|, |y₁|,|x₂|, |y₂| <= 1000.0).
    Please notice that A(x₁,y₁) and B(x₂,y₂)and C(x₃,y₃) are in an anticlockwise direction fromthe equilateral triangle. Andcoordinates A(x₁,y₁) and B(x₂,y₂)are given by anticlockwise.

输出

   For each test case, you should output the coordinate of C(x₃,y₃),the result should be rounded to 2 decimal places in a line.

示例输入

4

-100.00 0.00 0.000.00

0.00 0.00 0.00100.00

0.00 0.00 100.00100.00

1.00 0.00 1.8660.50

示例输出

(-50.00,86.60)

(-86.60,50.00)

(-36.60,136.60)

(1.00,1.00)

提示

 

来源

2013年山东省第四届ACM大学生程序设计竞赛

示例程序

推广结论：对于任意两个不同点A和B，A绕B旋转θ角度后的坐标为：

(Δx*cosθ- Δy * sinθ+ xB, Δy*cosθ + Δx * sinθ+ yB ) 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdio>
#include<math.h>
using namespace std;
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        double x1,x2,y1,y2,x3,y3,x,y;
        scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
        x=x2-x1;
        y=y2-y1;
        x3=x/2-y*sqrt(3)/2+x1;
        y3=y/2+x*sqrt(3)/2+y1;
        printf("(%.2f,%.2f)
",x3,y3);
    }
    return 0;
}